Question

我有以下简单的法典,有一类,包括普通建筑商和复印件建筑商。

class largeObj
{
public:
    largeObj()
    {
        printf("
正常建筑
");
    }

    largeObj(const largeObj& mv)
    {
        printf("
制模
");
    }

    ~largeObj()
    {
        printf("
拆除。
");
    }

    void tryme()
    {
        printf("
Hi :)
");
    }

};

largeObj iReturnLargeObjects()
{
    largeObj md;


    return md;
}


int main()
{

    largeObj mdd = iReturnLargeObjects();

    mdd.tryme();

    return 0;
}

The output is

正常建筑

制模

拆除。

iii

我也想到原因。

但是,如果我替代以下一线:

largeObj mdd = iReturnLargeObjects();

iii

largeObj& mdd = iReturnLargeObjects();

产出相同,原因何在?

I mean: shouldn t there be another copy in the first case (iiiout the &)? What s the difference between these two lines and why do they behave the same?

Answer 1

largeObj& mdd = iReturnLargeObjects();

你们不能把反常的“高价值”的提法束缚起来。这是非法的C++,只允许某些具体的汇编者延期。然而,即使这一提法是<条形码>,因此,转让是合法的,但问题的各个方面都没有变化。

您的产出没有不同,其原因就是编辑优化,称为“区域观察”。 C++中明确允许的这种优化标准允许汇编者在某些限制范围内跳出其认为没有必要的物体,即使这样做改变了方案的属人性,因此它是一种非常不寻常的优化。

底线是:在您的复印件/搬家和骑自行车上不产生副作用,因为即使贵方的节目正确性取决于他们的需要,汇编者也可以消除这些影响。

Answer 2

What compiler did you use, I m willing to bet this worked because whatever compiler you used didn t zero out unused stack variables after a return, you didn t get a seg fault because its still your stack. basically md was left on the stack which you are technically still allowed to address. The difference between the two is

大会 • 分娩;

它是整个变量的变量。

大会 Obj& mdd = iReturnLargeObjects();

这里指的是由于 la编纂者仍然存在的一个变量。

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